a parallel plate capacitor consisting of two square plates has a potential diffe

Question

a parallel plate capacitor consisting of two square plates has a potential difference V0 across the plates, and separated by a gap of 1.1mm. There is a vacuum between the plates. An electron real eased from rest at the surface of the negative plate crosses the capacitor and strikes the positive plate with a speed of
1.78*10^6 m/s. The charge on the electron is e= 1.6*10^-19 C and its
mass is me = 9.1*10^-31 kg.

What is the potential difference across the plates?

what is the capacitance of the device if each plate is a square, 1.45 cm on a side?

Explanation / Answer

a) We know that v^2 = u^2 + 2aS

a = v^2/2S = (1.78*10^6)^2/(2*1.1*10^-3) = 1.44*10^15 m/s^2

Now F = ma = qE = qV0/d =====> V0 = mad/q

V0 = (9.1*10^-31*1.44*10^15*1.1*10^-3)/(1.6*10^-19) = 9.009 ~ 9 V

b) A = (1.45*10^-2)^2 = 2.1*10^-4 m^2

C = e0*A/d = (8.85*10^-12*2.1*10^-4)/1.1*10^-3 = 1.7*10^-12 F = 1.7 pF

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